Let $P$ be a point in a compact Riemann surface $M$ of positive genus $g>0$. Then there does not exist a holomorphic function $f$ on $M\setminus\{P\}$ with a simple pole at $P$.
The statement is written like a trivial fact in Farkas & Kra Riemann surface textbook with a simple explanation: Since $g>0$.
I think it's from Riemann-Roch theorem (what else I can do?). So we consider $P$ as a divisor on $M$. Then the statement is equivalent to saying that $r(P^{-1}) = r(1) = 1$. From Riemann-Roch,
$$r(P^{-1}) = \deg P - g+1+i(P) = 2-g+i(P).$$
So the statement is equivalent to saying that $i(P) =0$, i.e., there is no holomorphic differential with degree $\geq 1$ at $P$. I can't see why this is true.